# Epsilon Delta Proof

1. Jun 1, 2008

### razored

[SOLVED] Epsilon Delta Proof

Does this limit proof make total sense? Given : "Show that $$\lim_{x \rightarrow 2} x^{2} = 4$$."

My attempt at it :$$0<|x^{2}-4|<\epsilon$$ which can also be written as $$0<|(x-2)(x+2)|<\epsilon$$.
$$0<|x-2|<\delta$$ where $$\delta > 0$$. It appears that $$\delta = \frac {\epsilon}{x+2}$$ which is the conversion factor. Which then by substitution, $$0<|x-2|<\frac {\epsilon}{x+2}$$.

Here is the actual proof:
Choose $$\delta = \frac {\epsilon}{x+2}$$, given $$epsilon > 0$$ then if $$0<|x-2|<\frac {\epsilon}{x+2}$$ then $$0<|(x-2)(x+2)|<\delta$$.

Is this explanation coherent? I actually have a slight idea of what I wrote. Hopefully I am on the right path.

Last edited: Jun 1, 2008
2. Jun 1, 2008

### dynamicsolo

One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

min ( 1 , $$\frac {\epsilon}{x+2}$$ ) ,

since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...

3. Jun 1, 2008

### razored

I can't believe I forgot to mention that!

4. Jun 1, 2008

### konthelion

When you are doing $$\epsilon-\delta$$ proofs, you have to say define what epsilon and delta are i.e. $$\forall \epsilon >0, \exists \delta$$ such that etc.

5. Jun 1, 2008

### eok20

Your $\delta$ should not depend on $x$. As the post above mentions, the logic goes: for all $\epsilon > 0$ there exists $\delta > 0$ such that if $|x-2| < \delta$ then $|x^2 - 4| < \epsilon$. Delta can, however, depend on epsilon. If |x-2| is bounded by $\delta < 1$, then what is a bound on |x+2|?

6. Jun 1, 2008

### dynamicsolo

Sorry -- quite so! It's been a while since I've looked at one of these proofs and I'm writing from someplace I don't have my books handy...

7. Jun 2, 2008

### razored

My second attempt : Given any $$\epsilon > 0$$, choose $$\delta = \frac {\epsilon} {|x+2|} (x\neq -2)$$ where $$\delta > 0$$, then $$|x-2||x+2|<\epsilon$$ whenever $$|x-2|<\delta = \frac {\epsilon} {|x+2|}$$.

Is this right?

Last edited: Jun 2, 2008
8. Jun 2, 2008

### razored

In my revision, I've been told that I did not prove anything. Is this a complete proof?

9. Jun 2, 2008

### konthelion

Let $$f(x)=x^2$$, $$x_{0}=2$$. $$\forall \epsilon>0,\exists \delta>0$$ such that $$|x^2-4|<\epsilon$$, $$|x-2|<\delta$$
Let $$\delta = \frac{\epsilon}{x+2}$$
... continue from here. "..." $$<\epsilon$$

OR let $$\delta=min\{ 1,\frac{\epsilon}{5} \}$$ and use the triangle inequality.
QED

Last edited: Jun 2, 2008
10. Jun 2, 2008

### razored

Substituting $$\delta = \frac {\epsilon}{x+2}$$ into $$|x-2|<\delta$$, we see that.. it works out? I do not know what to do from here.

Nor do I know the min way.

11. Jun 2, 2008

### konthelion

Continue from the proof,
Let $$\epsilon>0$$, let $$\delta = min\{\frac{\epsilon}{5}\}$$

Assume that $$|x-2|<\delta$$

By Triangle inequality,
$$|x^2-4|=|(x+2)(x-2)|\leq |x+2||x-2|$$

Since $$|x-2|<\delta$$, then $$|x-2|<\epsilon/5$$

Then for 0<x<3 such that $$|x+2|=5$$ **note: this is why you get the 5 as the denominator under epsilon.

Thus, $$|x^2-4|\leq |x+2||x-2|<5*\frac{\epsilon}{5}=\epsilon$$ QED

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